Ryan has 3 red lava lamps and 3 blue lava lamps. He arranges them in a row on a shelf randomly, and then randomly turns 3 of them on. What is the probability that the leftmost lamp is blue and off, and the rightmost lamp is red and on?
+6248 \(\text{Let's say we have an arrangement of lamps such that}\\ \text{the leftmost lamp is blue}\\ \text{and the rightmost lamp is red}\\ \text{Now we select 3 lamps at random and turn them on}\\ \text{There are a total of }\dbinom{6}{3}=20 \text{ ways they can be chosen}\\ \text{If we insist on the left lamp being off and the right one being on}\\ \text{we have }\dbinom{4}{2}=6 \text{ ways of selecting the remaining lamps that will go on}\\ \text{So given an arrangement as described above}\\ P[\text{choosing random 3 lamps to turn on results in left off and right on}]=\dfrac{6}{20}=\dfrac{3}{10}\)
\(\text{Now we need to determine the probability of}\\ \text{having the left lamp blue and right lamp red}\\ \text{We fix those two and for the remaining slots we have}\\ \dbinom{4}{2} = 6 \text{ way to select slots for say the red lamps}\\ \text{and having done that the blue lamps are determined}\\ \text{so there are 6 total arrangements satisfying the requirement}\\ \text{out of a possible }\dbinom{6}{3}=20 \text{ ways to arrange the lamps, so}\\ P[\text{leftmost lamp is blue, rightmost lamp is red}]=\dfrac{6}{20}=\dfrac{3}{10}\)
\(\text{The probability we're after is just the product of these two}\\ P[\text{left blue lamp off and right red lamp on}]=\left(\dfrac{3}{10}\right)^2=\dfrac{9}{100}\)
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